What is the time complexity of Longest Palindromic Subsequence After at Most K Operations?

The optimal solution for Longest Palindromic Subsequence After at Most K Operations runs in O(n^2 * k) time and uses O(n^2 * k) space. Quadratic in terms of string length and linear in terms of operations due to the 3D DP table.

What is the brute force solution for Longest Palindromic Subsequence After at Most K Operations?

The brute force approach for Longest Palindromic Subsequence After at Most K Operations is Brute Force, with O(2^n) time complexity and O(n) space complexity. Generate all possible subsequences and check if they can be transformed into a palindrome using at most k operations. This approach is inefficient due to the exponential growth of subsequences. The optimal Optimal Solution approach improves this to O(n^2 * k).

What algorithm or data structure is used to solve Longest Palindromic Subsequence After at Most K Operations?

Longest Palindromic Subsequence After at Most K Operations uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Two Pointers.

What are the interview tips for Longest Palindromic Subsequence After at Most K Operations?

Clarify the problem constraints and operations. Discuss potential edge cases upfront. Explain your thought process while coding.

What are common mistakes when solving Longest Palindromic Subsequence After at Most K Operations?

Overlooking the cost of character replacements. Not properly initializing the DP table.

#3472

Longest Palindromic Subsequence After at Most K Operations

Medium

StringDynamic ProgrammingDynamic ProgrammingTwo Pointers

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n)	O(n^2 * k)
Space	O(n)	O(n^2 * k)

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Intuition

Time O(n^2 * k)Space O(n^2 * k)

Use dynamic programming to build a table that tracks the longest palindromic subsequence lengths while considering the cost of character replacements.

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Algorithm

3 steps

1Step 1: Initialize a 3D DP array dp[i][j][k] to store lengths of palindromic subsequences from index i to j with k operations.
2Step 2: Fill the DP table based on character matches and allowable operations.
3Step 3: Return dp[0][n-1][k] for the final result.

solution.py11 lines

1def longest_palindromic_subseq(s, k):
2    n = len(s)
3    dp = [[[0] * (k + 1) for _ in range(n)] for _ in range(n)]
4    for length in range(1, n + 1):
5        for i in range(n - length + 1):
6            j = i + length - 1
7            if s[i] == s[j]:
8                dp[i][j][0] = dp[i + 1][j - 1][0] + 2
9            for ops in range(1, k + 1):
10                dp[i][j][ops] = max(dp[i][j][ops], dp[i + 1][j][ops - 1], dp[i][j - 1][ops - 1])
11    return dp[0][n - 1][k]

ℹ

Complexity note: Quadratic in terms of string length and linear in terms of operations due to the 3D DP table.

1Character replacements can help form palindromes.
2Dynamic programming efficiently tracks subsequence lengths.

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